At 22 °c an excess amount of a generic metal hydroxide m(oh)2 is mixed with pure water. the resulting equilibrium solution has a ph of 10.30. what is the ksp of the salt at 22 °c?

The balanced reaction equation:
M(OH)2 ↔ M^2+ + 2(OH)^-

and when the Ksp = [M^2+][OH-]^2
when PH + POH = 14
∴ POH = 14-10.3= 3.7 
and when POH = - ㏒ [OH-]
 3.7 = -㏒[OH-]
∴[OH] = 2x10^-4 
and when [M^2+] = 1/2[OH-]
∴[M^2+] = (2x10^-4) / 2 = 0.0001 M
So, by substitution in Ksp formula:
∴Ksp = (0.0001 * (2x10^-4)^2 = 4x10^-12

Sorry i dont understand the question can you rephase it?

Anyone know the answer?